Express sin^6theta in multiples of costheta and hence evaluate the integral of sin^6theta from 0 to pi/2.

Question

Thanks in advance! :)

Answer 1

sin^6θ = (sin^2θ)^3
= (1 - cos^2θ)^3
= 1 - 3cos^2θ + 3cos^4θ + cos^6θ

Now you can use your half-angle formula

cos^2θ = (1 + cos 2θ)/2

to get no exponents and multiples of θ.

You will end up with

1/192 (60θ - 45sin2θ + 9sin4θ - sin6θ)

from 0 to pi/2 yields 30pi/192